This is QFT, not QM. φ here is the operator (not a wavefuntion, like in QM), as is the canonical momentum.

I am not sure if we can carry the QM result that the commutator is i times the PB over to QFT. I was thinking more along the lines that the PB is explicitly given by {A,B}=dA/dφ(x) *dB/dπ(y) - dB/dφ(y) *dA/dπ(x).

If we work it out this way, you'll see that the PB in question reduces to ∂∂φ/∂φ

Humanino: This isnt QM - It is Quantum Field Theory, where we promote fields themselves to operators.

Everything you write is classical field theory, from my point of view. BTW, it's all very well explained by Susskind in a 2h or so lecture available freely, for instance if you're interested. It's an extremely important aspect of classical mechanics to be aware of before embarking on quantum field theory.NhNBW8a8-lI[/youtube]

If you worked out [tex]\{\Pi,\underline{\partial}\phi\}[/tex] you should be able to work this one with
[tex]\{f,g\}=-\{g,f\}[/tex]
[tex]\{f_1f_2,g\}=f_1\{f_2,g\}+f_2\{f_1,g\}[/tex]
As in
[tex]\{\Pi,\left(\underline{\partial}\phi\right)^2\} = \{\Pi,\underline{\partial}\phi\underline{\partial}\phi\} = \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\} + \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\} = 2 \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\}[/tex]

How about the result of [tex]\{\Pi,\underline{\partial}\phi\}[/tex] ? It seems to me there are several ways. I have been wondering, was the previous question about [tex]\{\Pi,\phi\}[/tex] by any chance ?
[tex]\{\Pi,\phi\}=\delta[/tex]

If you worked out [tex]\{\Pi,\underline{\partial}\phi\}[/tex] you should be able to work this one with
[tex]\{f,g\}=-\{g,f\}[/tex]
[tex]\{f_1f_2,g\}=f_1\{f_2,g\}+f_2\{f_1,g\}[/tex]
As in
[tex]\{\Pi,\left(\underline{\partial}\phi\right)^2\} = \{\Pi,\underline{\partial}\phi\underline{\partial}\phi\} = \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\} + \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\} = 2 \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\}[/tex]

Yes, I get this.

How about the result of [tex]\{\Pi,\underline{\partial}\phi\}[/tex] ? It seems to me there are several ways. I have been wondering, was the previous question about [tex]\{\Pi,\phi\}[/tex] by any chance ?
[tex]\{\Pi,\phi\}=\underline{\partial}\phi[/tex]

This is where I am stuck.

Because we're dealing with the Hamiltonian density, we have to work out the integral:

No this is wrong! the poisson bracket [itex]\{\pi, \partial_{t}\phi\}[/itex] is not zero! The field "velocity" [itex]\partial_{t}\phi[/itex] is a function of [itex]\pi , \nabla \phi[/itex] and [itex]\phi[/itex]